Topic 1108: Use of stacks
Time limit: 1 second
Memory Limit: 32 MB
Special Judgment: No
Submitted: 11671
Resolved: 3392
Topic description:
A stack is a basic data structure. The stack has two basic modes of operation, push and pop. Push pushes a value onto the top of the stack, while pop pops the value from the top of the stack. Now let's verify the use of the stack.
- enter:
-
For each set of test data, the first line is a positive integer n, 0<n<=10000 (n=0 ends). In the next n lines, the first character of each line may be 'P' or 'O' or 'A'; if it is 'P', it will be followed by an integer, which means to push the data onto the stack; if it is ' O' means pop the value on the top of the stack, if there is no element in the stack, ignore this operation; if it is 'A', it means to ask the value of the current top of the stack, if the stack is empty at that time, output 'E'. The stack starts out empty.
- output:
-
For each set of test data, process the stack according to the command characters in it; for all 'A' operations, output the value at the top of the stack at that time, each occupying a line, and output 'E' if the stack is empty at that time. When each set of test data is complete, output a blank line.
- Sample input:
-
3 A P 5 A 4 P 3 P 6 O A 0
- Sample output:
-
E 5 3
#include<stdio.h>
#include<stack>
#include<cstring>
using namespace std;
int main(){
int n;
stack<int> ch;
while(scanf("%d",&n)!=EOF&&n!=0){
// getchar();
while(!ch.empty()) ch.pop();
while(n--!=0){
char c[4];
scanf("%s",c);//Gets was originally used to read strings. The difference between it and scanf is that it can accept tabs such as spaces, TAB, etc.
//If you use gets, the numbers that need to be pushed on the stack by the P operation will also be read into a string, which is not as convenient as using scanf directly.
if(c[0]=='A'){
// if(ch.empty()){
// printf("go1\n");}
// if(!ch.empty()){
// printf("go2\n");}
if(!ch.empty()){
// printf("go!\n");
printf("%d\n",ch.top());
}
else
printf("E\n");
}
else if(c[0]=='O'){
if(!ch.empty()){
ch.pop();
}
}
else if(c[0]=='P'){
int a;
scanf("%d",&a);
// if(c[2]=='5')
// printf("ding!\n");
// else
// printf("%c\n",c[2]);
ch.push(a);
}
}
printf("\n");
}
return 0;
}
Topic 1101: Computational Expressions
Time limit: 1 second
Memory Limit: 32 MB
Special Judgment: No
Submitted: 7388
Resolved: 2307
- Topic description:
-
Evaluate an expression without parentheses
- enter:
-
There are many kinds of data, each group of data has one line, and there are no spaces in the expression
- output:
-
output result
- Sample input:
-
6/2+3+3*4
- Sample output:
-
18
#include<stack>
#include<cstdio>
#include<iostream>
using namespace std;
stack<int> op;
stack<double> in;
int mat[][5]={
1,0,0,0,0,
1,0,0,0,0,
1,0,0,0,0,
1,1,1,0,0,
1,1,1,0,0,
};
char str[1000];
void getE(double &retn,int &i,bool &reto){
if(i==0&&op.empty()==true){
reto = true;
retn = 0;
return;
}
if(str[i]==0){
reto = true;
retn = 0;
return;
}
if(str[i]>='0'&&str[i]<='9'){
straight=false;
}
else{
reto = true;
if(str[i]=='+')
retn = 1;
if(str[i]=='-')
retn = 2;
if(str[i]=='*')
retn = 3;
if(str[i]=='/')
retn = 4;
i++;
return;
}
retn = 0;
for(;str[i]<='9'&&str[i]>='0'&&str[i]!=0;i++){
retn * = 10;
retn + = str [i] - '0';
}
double d=1,dd=0;
if(str[i]=='.'){
i++;
for(;str[i]<='9'&&str[i]>='0';i++){
d/=10;
dd+=(str[i]-'0')*d;
}
}
retn + = dd;
return;
}
int main(){
while(scanf("%s",str)!=EOF){
while (! op.empty ()) op.pop ();
while(!in.empty()) in.pop();
double ren;
int index=0;
bool reop;
while(true){
getE(ren,index,reop);
if(reop==false){
in.push(ren);
}
else{
if (op.empty () == true || mat [(int) ren] [op.top ()] == 1)
op.push(ren);
else{
while(mat[(int)ren][op.top()]==0){
int o;
o=op.top();
op.pop ();
double b=in.top();
in.pop();
double a=in.top();
in.pop();
if(o==1)
a=a+b;
else if(o==2)
a=a-b;
else if(o==3)
a=a*b;
else if(o==4)
a=a/b;
in.push(a);
}
op.push(ren);
}
}
if(op.size()==2&&op.top()==0)
break;
}
cout<<in.top()<<endl;
}
return 0;
}
//1+3/2-1*3